A car enters a circular flat curve with a radius of curvature of 0.40 km with a maximum speed of 130 km/h. If the friction between the road and the car's tires can supply a centripetal acceleration, what is the static coefficient of friction?
R = 400 meters
v = 130,000/3600 = 36.1 m/s
Ac = v^2/R
= 3.26 m/s^2
Ac/g = mu = 3.26/9.81 = .322
To find the static coefficient of friction, we need to combine the information about the maximum speed and the radius of curvature.
The centripetal force required to keep the car moving in a circular path can be calculated using the formula:
F = m * a
where F is the centripetal force, m is the mass of the car, and a is the centripetal acceleration.
The centripetal force is provided by the friction force between the tires and the road:
F = µ * N
where µ is the coefficient of friction and N is the normal force.
The normal force can be calculated as:
N = m * g
where g is the acceleration due to gravity.
Since the centripetal acceleration is given and the mass of the car cancels out, we can solve for the coefficient of friction:
a = (v^2) / r
F = m * a = m * (v^2) / r
F = µ * N = µ * m * g
Set the two expressions for F equal to each other:
m * (v^2) / r = µ * m * g
Cancel out the mass:
(v^2) / r = µ * g
Solve for µ:
µ = (v^2) / (r * g)
Now, we can substitute the given values:
v = 130 km/h = 36.1 m/s
r = 0.40 km = 400 m
g ≈ 9.8 m/s^2
µ = (36.1^2) / (400 * 9.8)
µ ≈ 0.33
Therefore, the static coefficient of friction between the car's tires and the road is approximately 0.33.